Inductance of Coax Cable
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INDUCTANCE PER UNIT LENGTH FOR COAX CABLE Well, there is a well known formula for the self-inductance per unit length of a coaxial cable and it is (μ/2π) * ln(b/a) And I will show you how to obtain it: The definition of the self-inductance is phi = L * i L: self-inductance i: the current phi : the total magnetic flux passing through a surface that is orthogonal to the direction of the magnetic field in the coaxial cable. phi = B.dS ∫∫ But first we should find the expression of the magnetic induction field B inside the cable. For so, we use Ampere's theorem : the curl of B on an adequately chosen contour equals μ*i Therefore B.dl = μ*i ∫ Now, by examining the symmetry of the coaxial cable we notice that the magnetic field lines are concentric circles around the cable , that means that the vector B has only one component along the vector θ ( the second component in the cylindrincal coordinates system (ρ,θ,z).; and the magnitude of B is only function of the radius ρ. So we can write vectorB = B(ρ). vectorθ. Since B only depends of the distance ρ it would be smart to chose the contour as circle of radius ρ. Therefore B(ρ) will be constant on that contour and can be moved out of the contour. So (on the cirlce) B.dl = B(ρ) dl . ∫ ∫ dl is the circumference of the circle of radius ρ. dl=2πρ ∫ ∫ So B(ρ) * 2πρ = μ*i And we find the expression of B inside the cable (a<ρ<b): B(ρ) = μ*i / (2πρ) Now, we still have to calculate the total magnetic flux: phi = B.dS. ∫∫ To calculate this surface integral we should find the radiated surface by the magnetic field. The ideal surface would be a vertical rectangle of height deltaZ included in a plane containing the axis of the cable. So the elementry surface dS=dρdz phi = B.dS = (a<ρ<b) (0<z<deltaZ) [ μ*i / (2πρ) dρdz ] ∫∫ ∫ ∫
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Inductance per unit length of coax cables
Transcript of Inductance of Coax Cable
INDUCTANCE PER UNIT LENGTH FOR COAX CABLEWell, there is a well known formula for the self-inductance per unit length of a coaxial cable and it is(/2) * ln(b/a)
And I will show you how to obtain it:The definition of the self-inductance is phi = L * iL: self-inductancei: the currentphi : the total magnetic flux passing through a surface that is orthogonal to the direction of the magnetic field in the coaxial cable.phi = B.dS
But first we should find the expression of the magnetic induction field B inside the cable.For so, we use Ampere's theorem : the curl of B on an adequately chosen contour equals *iTherefore B.dl = *i
Now, by examining the symmetry of the coaxial cable we notice that the magnetic field lines are concentric circles around the cable , that means that the vector B has only one component along the vector ( the second component in the cylindrincal coordinates system (,,z).; and the magnitude of B is only function of the radius .So we can writevectorB = B(). vector. Since B only depends of the distance it would be smart to chose the contour as circle of radius . Therefore B() will be constant on that contour and can be moved out of the contour.
So (on the cirlce) B.dl = B() dl .dl is the circumference of the circle of radius . dl=2So B() * 2 = *iAnd we find the expression of B inside the cable (a
